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Mar 1, 2018 at 23:28 vote accept Mendieta
Mar 1, 2018 at 13:07 history edited YCor
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Mar 1, 2018 at 8:21 answer added YCor timeline score: 5
Mar 1, 2018 at 3:46 comment added Mendieta @Mohan: Maybe your remark about local $\mathbb{C}$-algebras can be extended to an example showing that the answer to my question is 'no', but I think that the question is clear.
Mar 1, 2018 at 2:58 comment added Mohan You are right, but you did not my question about local rings.
Mar 1, 2018 at 2:21 comment added Johannes Hahn @Mohan: That's does not satisfy the condition $\forall g,h: B\to\mathbb{C}: g\circ f=h\circ f$. Polynomial rings won't work because the condition implies $im(f) \subseteq \mathbb{C} + rad(B)$.
Mar 1, 2018 at 1:53 comment added Mohan I have two questions. Local $\mathbb{C}$-algebras are rarely finitely generated, so they are not in $\mathcal{A}$, so the second part is not clear. Secondly, have you considered the map $A=\mathbb{C}[t^2,t^3]\subset \mathbb{C}[t]$ for your $f$?
Mar 1, 2018 at 0:01 review First posts
Mar 1, 2018 at 1:13
Feb 28, 2018 at 23:57 history asked Mendieta CC BY-SA 3.0